Single-phase half controlled bridge rectifier with RL load. Size of L

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Hello, I need to dimension the inductance L so that the current ripple from the output current (approximatelly 30A) is not greater than 20 percent.
Please look at the picture below. I don't quite know how to do this.
kind regards

 

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An exact solution of the exercise problem needs to consider the sine voltage waveform and solve the differential equation.

It's not clear if simplifications are permitted, one simplification is obviously the assumption of ideal SCR/diode characteristics because to forward voltage is specified.

 

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Yes Simplifications are permitted. The voltages of the thyristors and diodes are negligible.
The mean value of the cut sine wave output voltage may be used.
I just don't know which time Δt to use for the formula u_L = L * Δi_pp / Δt.
 

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Hi,

R and L form a low pass filter.

Simplified if you want to attenuate the ripple to 20% which equals 1/5, then set the cutoff frequency of the filter also to 1/5 of your fundamental ripple frequency.

****

Btw: don´t call it "I_constant", or "constant current" better call it "average current". This also meets your Ohm´s law calculation.
You always have to use identical values, like: V_peak/I_peak ... V_avg/I_avg .. V_RMS/ I_RMS .. and so on.

Klaus
 

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fg=1/(2*pi*L*R) that is the formula for the cutoff frequency.

Fundamental ripple Period length is T/2 * 1/3 = T/6 . T/2 is because the inductor is charged twice per period and 1/3 is because the ignition angle of the thyristor is 120 degrees (i.e. the inductor is actually charged for 1/3 of half the period).
So the fundamental ripple frequency is 6/T = 6/20ms = 300Hz.

So the formula is 300/5 = 1/(2*pi*L*R)
-> L= 5/(2*pi*R*300) = 1.846mH. (R=1.4368)

Is this calculation correct?
 

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Obviously ripple frequency is 100 Hz (full bridge rectifier). As mentioned, an exact calculation need to refer to voltage waveform. Triangle approximation may be an appropriate simplification.

If you reviewed my simulation, you know that the correct solution is around 50 mH.
 

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Hi,

I just wondered why 50mH ... while my calculation gives about 5.5mH.

Indeed I thought my calculation is on the safe side.
Then I recognized that FvM calculated the 20% from the actual average.
But my approach calculates the 20% with respect to I_PP of circuit without inductor.
The I_PP without inductor is almost 200A.

After re reading post#1 I assume FvM´s aproach is the correct one. (Referencing to I_avg)

Sorry for the confusion. Ignore my post#4

Klaus
 

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The 50mH is correct. I simulated it with Plexim.
Is there any way I can quickly calculate this value (only approximately) (without simulation and without exact calculations). So that when I simulate a circuit I have an approximate guide value of how big it needs to be.

I have tried it like this:

L= (Vout,average * T/4)/ΔiL = ( (51.7253V* T/4) / (7.2A) = 36mH.

T/4 because the current rises twice and falls twice per period. So one rise takes T/4.
Vout,average is simply the average value of the output voltage and ΔiL is the permissible ripple current.

Is there still room for improvement here, perhaps a different voltage and not the average value of the output voltage? It should only be a first approximation. Thank you
 

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Years ago a choke filter was more frequently suggested to smooth waveforms coming from the diode bridge. Your schematic is similar. (The choke filter has a different effect from capacitive smoothing.) The choke is selected to block the waveform's frequency yet to present very little impedance to DC. Suppose you want the choke to present an additional 10 times your load impedance at 100 Hz. Although the waveform isn't a sine, this is just an approximation. Then the choke should present 14 ohms impedance to 100 Hz.
Formula for inductive impedance:
2πfL,
or:
14=6*100*?
To solve for the question mark substitute 23 mH.
 

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You get a first order estimation of L by assuming linear current decrease. Determine current fall time tf = 10 ms*(arcsin(1.1*51.7/325) + 120°)/180° = 7.23 ms.
L = tf*51.7/7.2 = 51.9 mH. The deviation from 50 mH is caused by abstracting from actual exponential current waveform.
 

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But why the fall time and not the rise time? Because the inductance charges when voltage is applied.
 

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FvM said:
Voltage integral during diode conduction phase is easier to calculate due to constant voltage. SCR conduction phase involves sin integral etc.
Click to expand...
Okey, but you can still take the average value of the output voltage, because the coil has charged to approximately this average value during the charging phase, so it will also have this value when discharging (approximately)? That's the idea behind it, isn't it?
 

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